Metamath Proof Explorer


Theorem nnncan

Description: Cancellation law for subtraction. (Contributed by NM, 4-Sep-2005)

Ref Expression
Assertion nnncan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) − 𝐶 ) = ( 𝐴𝐵 ) )

Proof

Step Hyp Ref Expression
1 subcl ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
2 1 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
3 subsub4 ( ( 𝐴 ∈ ℂ ∧ ( 𝐵𝐶 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) − 𝐶 ) = ( 𝐴 − ( ( 𝐵𝐶 ) + 𝐶 ) ) )
4 2 3 syld3an2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) − 𝐶 ) = ( 𝐴 − ( ( 𝐵𝐶 ) + 𝐶 ) ) )
5 npcan ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐵 )
6 5 oveq2d ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( ( 𝐵𝐶 ) + 𝐶 ) ) = ( 𝐴𝐵 ) )
7 6 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( ( 𝐵𝐶 ) + 𝐶 ) ) = ( 𝐴𝐵 ) )
8 4 7 eqtrd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) − 𝐶 ) = ( 𝐴𝐵 ) )